Answer:
Step-by-step explanation:
When using the substitution method we use the fact that if two expressions y and x are of equal value x=y, then x may replace y or vice versa in another expression without changing the value of the expression.
Solve the systems of equations using the substitution method
{y=2x+4
{y=3x+2
We substitute the y in the top equation with the expression for the second equation:
2x+4 = 3x+2
4−2 = 3x−2
2=== = x
To determine the y-value, we may proceed by inserting our x-value in any of the equations. We select the first equation:
y= 2x + 4
We plug in x=2 and get
y= 2⋅2+4 = 8
The elimination method requires us to add or subtract the equations in order to eliminate either x or y, often one may not proceed with the addition directly without first multiplying either the first or second equation by some value.
Example:
2x−2y = 8
x+y = 1
We now wish to add the two equations but it will not result in either x or y being eliminated. Therefore we must multiply the second equation by 2 on both sides and get:
2x−2y = 8
2x+2y = 2
Now we attempt to add our system of equations. We commence with the x-terms on the left, and the y-terms thereafter and finally with the numbers on the right side:
(2x+2x) + (−2y+2y) = 8+2
The y-terms have now been eliminated and we now have an equation with only one variable:
4x = 10
x= 10/4 =2.5
Thereafter, in order to determine the y-value we insert x=2.5 in one of the equations. We select the first:
2⋅2.5−2y = 8
5−8 = 2y
−3 =2y
−3/2 =y
y =-1.5
Answer:
hi
Step-by-step explanation:
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
Answer: 
<u>Step-by-step explanation:</u>
Convert everything to "sin" and "cos" and then cancel out the common factors.
![\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)](https://tex.z-dn.net/?f=%5Cdfrac%7Bcot%28x%29%2Bcsc%28x%29%7D%7Bsin%28x%29%2Btan%28x%29%7D%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%28%5Cdfrac%7Bsin%28x%29%7D%7B1%7D%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%5B%5Cdfrac%7Bsin%28x%29%7D%7B1%7D%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%5D%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%28%5Cdfrac%7Bsin%28x%29cos%28x%29%7D%7Bcos%28x%29%7D%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29)
Step-by-step explanation:
sin theta = opp / hyp
sin theta × hyp = opp
sin60 × 30 = opp
.866 × 30 = opp
25.98° = opp
*Theta is the 0 with a line through it
